What is the integral of (x-29)/((x-4)(x+1)) ?

The integral of (x-29)/((x-4)(x+1)) is -5ln|x-4|+6ln|x+1|+C

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integral of (x-29)/((x-4)(x+1))

\int \frac{x - 29}{( x - 4 ) ( x + 1 )} d x

- 5 ln ∣ x - 4 ∣ + 6 ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{x - 29}{( x - 4 ) ( x + 1 )} d x

Take the partial fraction of

\frac{x - 29}{( x - 4 ) ( x + 1 )} : - \frac{5}{x - 4} + \frac{6}{x + 1}

= \int - \frac{5}{x - 4} + \frac{6}{x + 1} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{5}{x - 4} d x + \int \frac{6}{x + 1} d x

\int \frac{5}{x - 4} d x = 5 ln ∣ x - 4 ∣

\int \frac{6}{x + 1} d x = 6 ln ∣ x + 1 ∣

= - 5 ln ∣ x - 4 ∣ + 6 ln ∣ x + 1 ∣

Add a constant to the solution

= - 5 ln ∣ x - 4 ∣ + 6 ln ∣ x + 1 ∣ + C

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What is the integral of (x-29)/((x-4)(x+1)) ?
The integral of (x-29)/((x-4)(x+1)) is -5ln|x-4|+6ln|x+1|+C